A 0.50-kilogram puck sliding on a horizontal

shuffleboard court is slowed to rest by a

frictional force of 1.2 newtons. What is the

coefficient of kinetic friction between the puck

and the surface of the shuffleboard court?

(1) 0.24 (3) 0.60

(2) 0.42 (4) 4.1

Question

Jerry

Physics

25 December, 19:25

A 0.50-kilogram puck sliding on a horizontal

shuffleboard court is slowed to rest by a

frictional force of 1.2 newtons. What is the

coefficient of kinetic friction between the puck

and the surface of the shuffleboard court?

(1) 0.24 (3) 0.60

(2) 0.42 (4) 4.1

+5

Answers (1)

Know the Answer?

Giancarlo Garcia · Answer 1

-The u is what you are trying to find, also known as coefficient of friction.

-Fn is your mass in Newtons, which is 0.5 x 9.8 which equals 4.9

-Fmax is your frictional force.

When you solve for, your equation becomes

/ Fn =

= 1.2N/4.9N

(coefficient of friction) = 0.25

A 0.50-kilogram puck sliding on a horizontalshuffleboard court is slowed to rest by africtional force of 1.2 newtons. What is thecoefficient of kinetic friction between the puckand the surface of the shuffleboard court?(1) 0.24 (3) 0.60(2) 0.42 (4) 4.1

A 0.50-kilogram puck sliding on a horizontal

shuffleboard court is slowed to rest by a

frictional force of 1.2 newtons. What is the

coefficient of kinetic friction between the puck

and the surface of the shuffleboard court?

(1) 0.24 (3) 0.60

(2) 0.42 (4) 4.1